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Minimize count of flips required to make sum of the given array equal to 0
  • Last Updated : 05 Feb, 2021

Given an array arr[] consisting of N integers, the task is to minimize the count of elements required to be multiplied by -1 such that the sum of array elements is 0. If it is not possible to make the sum 0, print “-1”.
 

Examples:

 

Input: arr[] = {2, 3, 1, 4}
Output: 2
Explanation: 
Multiply arr[0] by -1. Therefore, the array modifies to {-2, 3, 1, 4}. 
Multiply arr[1] by -1. Therefore, the array modifies to {-2, -3, 1, 4} 
Therefore, the sum of the modified array is 0 and the minimum operations required is 2.

Input: arr[] = {2}
Output: -1.



 

 

Naive Approach: The simplest approach is to divide the array into two subsets in every possible way. For each division, check if the difference of their subset-sum is 0 or not. If found to be 0, then the length of the smaller subset is the result.
Time Complexity: O(2N)
Auxiliary Space: O(N)

 

Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming. Follow the steps to solve the problem:

 

  • To make the sum of all array elements equal to 0, divide the given array elements into two subsets having an equal sum.
  • Out of all the subsets possible of the given array, the subset whose size is the minimum of all is chosen.
  • If the sum of the given array is odd, no subset is possible to make the sum 0, hence return -1
  • Else, try all possible subset sums of the array and check if the sum of the subset is equal to sum/2. where the sum is the sum of all elements of the array.
  • The recurrence relation of dp[] is:

dp(i, j) = min (dp(i+1, j – arr[i]]+1), dp(i+1, j))
where 
 

dp (i, j) represents the minimum operations to make sum j equal to 0 using elements having index [i, N-1].
j represents the current sum.
i represents the current index.

 

  • Using the above recurrence, print dp(0, sum/2) as the result.

 

Below is the implementation of the above approach:

 

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Initialze dp[][]
int dp[2001][2001];
 
// Function to find the nimimum number
// of operations to make sum of A[] 0
int solve(vector<int>& A, int i,
          int sum, int N)
{
    // Initialize answer
    int res = 2001;
 
    // Base case
    if (sum < 0 or (i == N and sum != 0)) {
        return 2001;
    }
 
    // Otherwise, return 0
    if (sum == 0 or i >= N) {
        return dp[i][sum] = 0;
    }
 
    // Pre-computed subproblem
    if (dp[i][sum] != -1) {
        return dp[i][sum];
    }
 
    // Recurrence relation for finding
    // the minimum of the sum of subsets
    res = min(solve(A, i + 1, sum - A[i], N) + 1,
              solve(A, i + 1, sum, N));
 
    // Return the result
    return dp[i][sum] = res;
}
 
// Function to find the minimum number
// of elements required to be flipped
// to amke sum the array equal to 0
void minOp(vector<int>& A, int N)
{
    int sum = 0;
 
    // Find the sum of array
    for (auto it : A) {
        sum += it;
    }
 
    if (sum % 2 == 0) {
 
        // Initialise dp[][]  with -1
        memset(dp, -1, sizeof(dp));
 
        int ans = solve(A, 0, sum / 2, N);
 
        // No solution exists
        if (ans < 0 || ans > N) {
            cout << "-1" << endl;
        }
 
        // Otherwise
        else {
            cout << ans << endl;
        }
    }
 
    // If sum is odd, no
    // subset is possible
    else {
        cout << "-1" << endl;
    }
}
 
// Driver Code
int main()
{
    vector<int> A = { 2, 3, 1, 4 };
    int N = A.size();
 
    // Function Call
    minOp(A, N);
 
    return 0;
}

Java




// Java program for the above approach
class GFG
{
 
// Initialze dp[][]
static int [][]dp = new int[2001][2001];
 
// Function to find the nimimum number
// of operations to make sum of A[] 0
static int solve(int []A, int i,
          int sum, int N)
{
   
    // Initialize answer
    int res = 2001;
 
    // Base case
    if (sum < 0 || (i == N && sum != 0))
    {
        return 2001;
    }
 
    // Otherwise, return 0
    if (sum == 0 || i >= N)
    {
        return dp[i][sum] = 0;
    }
 
    // Pre-computed subproblem
    if (dp[i][sum] != -1)
    {
        return dp[i][sum];
    }
 
    // Recurrence relation for finding
    // the minimum of the sum of subsets
    res = Math.min(solve(A, i + 1, sum - A[i], N) + 1,
              solve(A, i + 1, sum, N));
 
    // Return the result
    return dp[i][sum] = res;
}
 
// Function to find the minimum number
// of elements required to be flipped
// to amke sum the array equal to 0
static void minOp(int []A, int N)
{
    int sum = 0;
 
    // Find the sum of array
    for (int it : A)
    {
        sum += it;
    }
 
    if (sum % 2 == 0)
    {
 
        // Initialise dp[][]  with -1
        for(int i = 0; i < 2001; i++)
        {
            for (int j = 0; j < 2001; j++)
            {
                dp[i][j] = -1;
            }
        }
 
        int ans = solve(A, 0, sum / 2, N);
 
        // No solution exists
        if (ans < 0 || ans > N)
        {
            System.out.print("-1" +"\n");
        }
 
        // Otherwise
        else
        {
            System.out.print(ans +"\n");
        }
    }
 
    // If sum is odd, no
    // subset is possible
    else
    {
        System.out.print("-1" +"\n");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int []A = { 2, 3, 1, 4 };
    int N = A.length;
 
    // Function Call
    minOp(A, N);
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python program for the above approach
 
# Initialze dp[][]
dp = [[-1 for i in range(2001)] for j in range(2001)]
 
# Function to find the nimimum number
# of operations to make sum of A[] 0
def solve(A, i, sum, N):
   
    # Initialize answer
    res = 2001
 
    # Base case
    if (sum < 0 or (i == N and sum != 0)):
        return 2001
 
    # Otherwise, return 0
    if (sum == 0 or i >= N):
        dp[i][sum] = 0
        return 0
 
    # Pre-computed subproblem
    if (dp[i][sum] != -1):
        return dp[i][sum]
 
    # Recurrence relation for finding
    # the minimum of the sum of subsets
    res = min(solve(A, i + 1, sum - A[i], N) + 1,
              solve(A, i + 1, sum, N))
 
    # Return the result
    dp[i][sum] = res
    return res
 
# Function to find the minimum number
# of elements required to be flipped
# to amke sum the array equal to 0
def minOp(A, N):
    sum = 0
 
    # Find the sum of array
    for it in A:
        sum += it   
    if (sum % 2 == 0):
       
        # Initialise dp[][]  with -1
        dp = [[-1 for i in range(2001)] for j in range(2001)]
        ans = solve(A, 0, sum // 2, N)
 
        # No solution exists
        if (ans < 0 or ans > N):
            print("-1")
         
        # Otherwise
        else:
            print(ans)
 
    # If sum is odd, no
    # subset is possible
    else:
        print(-1)
 
# Driver Code
A = [ 2, 3, 1, 4 ]
N = len(A)
 
# Function Call
minOp(A, N)
 
# This code is contributed by rohitsingh07052.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG
{
 
// Initialze [,]dp
static int [,]dp = new int[2001,2001];
 
// Function to find the nimimum number
// of operations to make sum of []A 0
static int solve(int []A, int i,
          int sum, int N)
{
   
    // Initialize answer
    int res = 2001;
 
    // Base case
    if (sum < 0 || (i == N && sum != 0))
    {
        return 2001;
    }
 
    // Otherwise, return 0
    if (sum == 0 || i >= N)
    {
        return dp[i, sum] = 0;
    }
 
    // Pre-computed subproblem
    if (dp[i, sum] != -1)
    {
        return dp[i, sum];
    }
 
    // Recurrence relation for finding
    // the minimum of the sum of subsets
    res = Math.Min(solve(A, i + 1, sum - A[i], N) + 1,
              solve(A, i + 1, sum, N));
 
    // Return the result
    return dp[i, sum] = res;
}
 
// Function to find the minimum number
// of elements required to be flipped
// to amke sum the array equal to 0
static void minOp(int []A, int N)
{
    int sum = 0;
 
    // Find the sum of array
    foreach (int it in A)
    {
        sum += it;
    }
 
    if (sum % 2 == 0)
    {
 
        // Initialise [,]dp  with -1
        for(int i = 0; i < 2001; i++)
        {
            for (int j = 0; j < 2001; j++)
            {
                dp[i, j] = -1;
            }
        }
 
        int ans = solve(A, 0, sum / 2, N);
 
        // No solution exists
        if (ans < 0 || ans > N)
        {
            Console.Write("-1" +"\n");
        }
 
        // Otherwise
        else
        {
            Console.Write(ans +"\n");
        }
    }
 
    // If sum is odd, no
    // subset is possible
    else
    {
        Console.Write("-1" +"\n");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int []A = { 2, 3, 1, 4 };
    int N = A.Length;
 
    // Function Call
    minOp(A, N);
}
}
 
// This code is contributed by 29AjayKumar
Output: 
2

 

Time Complexity: O(S*N), where S is the sum of the given array.
Auxiliary Space: O(S*N)

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